Given that: a,b,c,d are positive real number, and the sum of them is 4. :Consider that ( chứng minh rằng)
\(\frac{1}{ab}+\frac{1}{cd}\ge\frac{a^2+b^2+c^2+d^2}{2}\)
Những câu hỏi liên quan
Bài thi số 3
19:25
Câu 1:
A man drove a car from A to B at speed 60km/h. After arriving B, he took a rest for 30 minutes then turned back to A at speed 40km/h. Known that he started from A at 7:00 am and he reached A again at 3:15pm on the same day. The distance between A and B is km.
Câu 2:
The minimum of the expression is
Câu 3:
Given that is a positive integer such that and are perfect squares.
The sum of such integers is
Câu 4:
Given two triangles and...
Đọc tiếp
Bài thi số 3
A man drove a car from A to B at speed 60km/h. After arriving B, he took a rest for 30 minutes then turned back to A at speed 40km/h. Known that he started from A at 7:00 am and he reached A again at 3:15pm on the same day. The distance between A and B is km. Câu 2:
The minimum of the expression(x+2)(x+3)(x+6)$)
is
Câu 3:
Given that
is a positive integer such that and 
are perfect squares.
The sum of such integers is
Câu 4:
Given two triangles
and 
. Known that 
, 
and 
.
If
then 
Câu 5:
How many real numbers are there such that 
?
Answer: There are numbers.
Câu 6:
The operation
on two numbers produces a number equal to their sum minus 2.The value of @3\)...@2017\)$)
is
Câu 7:
ABC is a triangle. AM is the bisector of angle CAB. Given that AM = 4cm, AB = 6m and AC = 12cm.Then the measurement of angle BAC is degrees. Câu 8:
In the equation above, where 
is a constant.The greatest possible value of such that the equation has at least one solution is
Câu 9:
and 
are positive integers such that =n^2+1$)
, where 
is a prime number.
The number of pairs$)
is
Câu 10:
Given that=\frac{x^2}{2x-2x^2-1}$)
.
Calculate:
+f\(\frac{2}{2016}\)+f\(\frac{3}{2016}\)+...+f\(\frac{2015}{2016}\)+f\(\frac{2016}{2016}\)$)
=
(Input the answer as a decimal in its simplest form) Nộp bài
19:25 Câu 1:
A man drove a car from A to B at speed 60km/h. After arriving B, he took a rest for 30 minutes then turned back to A at speed 40km/h. Known that he started from A at 7:00 am and he reached A again at 3:15pm on the same day. The distance between A and B is km. Câu 2:
The minimum of the expression
Given that
The sum of such integers
Given two triangles
If
How many real numbers
Answer: There are numbers
The operation
ABC is a triangle. AM is the bisector of angle CAB. Given that AM = 4cm, AB = 6m and AC = 12cm.Then the measurement of angle BAC is degrees. Câu 8:
The number of pairs
Given that
Calculate:
(Input the answer as a decimal in its simplest form) Nộp bài
câu 7 mk bấm nhầm đáp án là 120
qua B kẻ đường thẳng song song với AM cắt AC ở N.
vì AM là phân giác góc BAC nên có :
\(\dfrac{AC}{AB}=\dfrac{CM}{BM}=\dfrac{12}{6}=2\) suy ra \(\dfrac{CM}{BC}=\dfrac{CM}{CM+BM}=\dfrac{12}{12+6}=\dfrac{2}{3}\)
vì AM song song với BN nên có :
1,\(\dfrac{CA}{AN}=\dfrac{CM}{BM}=\dfrac{12}{AN}=2\) suy ra AN=6
2,\(\dfrac{AM}{BN}=\dfrac{CM}{BC}=\dfrac{2}{3}=\dfrac{4}{BN}\)suy ra BN=6
vì AB=6 nên tam giác ABN đều
suy ra \(\widehat{NAB}\)=\(60^0\)
mà \(\widehat{NAB}+\widehat{BAC}=\)\(180^0\)
nên \(\widehat{BAC}=\)\(120^0\)
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Câu1 : 186
Câu2 :-36
Câu3 : 19
Câu4 : 20
Câu5 : 0
Câu6 : 2017
câu7 : 110
Câu8 : 2
Câu9 : 2
Cau10 : 1
Mình lm được có 80 thui ko bt sai chỗ nào.
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Let a , b and c be positive real numbers such that a + b + c = 3 . Find the minimum value of the expression .
\(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2+b^2+c^2}\)
Cho \(a,b,c,d>0\). Chứng minh rằng: \(\frac{1}{a^2+ab}+\frac{1}{b^2+bc}+\frac{1}{c^2+cd}+\frac{1}{d^2+da}\ge\frac{4}{ac+bd}\)
Svacxo chăng :33 Ai thử đi, e sợ biến nhiều lắm :))
Let a,b be the roots of equation \(x^2-px+q=0\) and let c,d be the roots of the equation \(x^2-rx+s=0\), where p,q,r,s are some positive real numbers. Suppose that :
\(M=\frac{2\left(abc+bcd+cda+dab\right)}{p^2+q^2+r^2+s^2}\)
is an integer. Determine a,b,c,d .
1) ABC is a triangle where M is the midpoint of segment BC.
MD and ME are two bisectors of triangles AMB and AMC respectively.
If AM m; BC a . Then DE ???
2)dfrac{1}{left(x+29right)^2}+dfrac{1}{left(x+30right)^2}dfrac{5}{4}
What is the product of all real solutions to the equation above?
3) The sum of all possible natural numbers n such that
n^2+n+1589 is a perfect square is.....
4) Given that x is a positive integer such that x and x+99 are perfect squares
The sum of integer x is ....
Đọc tiếp
1) ABC is a triangle where M is the midpoint of segment BC.
MD and ME are two bisectors of triangles AMB and AMC respectively.
If AM= m; BC = a . Then DE = ???
2)\(\dfrac{1}{\left(x+29\right)^2}+\dfrac{1}{\left(x+30\right)^2}=\dfrac{5}{4}\)
What is the product of all real solutions to the equation above?
3) The sum of all possible natural numbers n such that
\(n^2+n+1589\) is a perfect square is.....
4) Given that x is a positive integer such that x and x+99 are perfect squares
The sum of integer x is ...
5)The operation @ on two numbers produces a number equal to their sum minus 2. The value of
(...((1@2)@3....@2017)
6) Given f(x)=\(\dfrac{x^2}{2x-2x^2-1}\)
=> \(f\left(\dfrac{1}{2016}\right)+f\left(\dfrac{2}{2016}\right)+f\left(\dfrac{3}{2016}\right)+...+f\left(\dfrac{2016}{2016}\right)\)
Các bn giúp mk vs >>> tks nha!!!
?????????????????????????????????????????????? Are you learning English or Math? I'm sure you are're mistake of English
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1. Find x, given that:
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{3}{40}\right)+\left(x+\frac{3}{88}\right)=\frac{31}{22}\)
2. The arithmatic average of three numbers is equal to 13.5 . The average of the first number and second number is 13, and the average of the second number and third number is 13.85 . Find the three numbers.
3. Compare A and B, given that:
A = 2.015 x 201.7
B = 20.16 x 20.16
Cho a,b,c,d>0 và a+b+c+d=4
Chúng minh rằng \(\frac{1}{ab}+\frac{1}{cd}\ge\frac{a^2+b^2+c^2+d^2}{2}\)
frac{a}{1+b^{2}c}+frac{b}{1+c^{2}d}+frac{c}{1+d^{2}a}+frac{d}{1+a^{2}b}geq 2$Ta có $sum frac{a}{1+b^2c}sum frac{a^2}{a+ab^2c}$Áp dụng Cauchy-Schwarzt ta có $sum frac{a}{1+b^2c}sum frac{a^2}{a+ab^2c}geq frac{(a+b+c+d)^2}{a+b+c+d+ab^2c+bc^2d+cd^2a+da^2b}frac{16}{4+ab^2c+bc^2d+cd^2a+da^2b}$Do đó ta chỉ cần chứng minh $ab^2c+bc^2d+cd^2a+da^2bleq 4$ là suy ra $sum frac{a}{1+b^2c}geq frac{16}{4+4}2$Bất đẳng thức đã cho tương đương $ab.bc+bc.cd+cd.da+da.ableq 4$ với $a+b+c+d4$Chuyển $l...
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\frac{a}{1+b^{2}c}+\frac{b}{1+c^{2}d}+\frac{c}{1+d^{2}a}+\frac{d}{1+a^{2}b}\geq 2$
Ta có $\sum \frac{a}{1+b^2c}=\sum \frac{a^2}{a+ab^2c}$
Áp dụng Cauchy-Schwarzt ta có
$\sum \frac{a}{1+b^2c}=\sum \frac{a^2}{a+ab^2c}\geq \frac{(a+b+c+d)^2}{a+b+c+d+ab^2c+bc^2d+cd^2a+da^2b}=\frac{16}{4+ab^2c+bc^2d+cd^2a+da^2b}$
Do đó ta chỉ cần chứng minh $ab^2c+bc^2d+cd^2a+da^2b\leq 4$ là suy ra $\sum \frac{a}{1+b^2c}\geq \frac{16}{4+4}=2$
Bất đẳng thức đã cho tương đương $ab.bc+bc.cd+cd.da+da.ab\leq 4$ với $a+b+c+d=4$
Chuyển $\left ( ab,bc,cd,da \right )\Rightarrow (x,y,z,t)$
Ta có $x+y+z+t=ab+bc+cd+ad \leq \frac{(a+b+c+d)^2}{4}=4$
Lại có $ab^2c+bc^2d+cd^2a+da^2b=xy+yz+zt+tx \leq \frac{(x+y+z+t)^2}{4} \leq \frac{4^2}{4}=4$
Vậy ta có đpcm
Dấu = xảy ra khi $a=b=c=d=1$
doc lam sao
Cho a,b,c,d>0 và a+b+c+d=4
Chúng minh rằng \(\frac{1}{ab}+\frac{1}{cd}\ge\frac{a^2+b^2+c^2+d^2}{2}\)
ta có ab( a\(^2\)+b\(^2\))\(\le\)2( tự CM)
=> ( a\(^2\)+ b\(^2\))\(\le\)2/ab
=> ( a\(^2\)+ b\(^2\))/2\(\le\)1/ab
làm tương tự ta có ( c\(^2\)+d\(^2\))/2\(\le\)1/cd
cộng vế tương ứng vế. Hết.
mình dùng tv ₫ể viết, có một Số chỗ hơi "khắm". Xin thứ lỗi.
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Bạn Huy Le ơi, cho mik hỏi tại sao ab(a^2+b^2)<=2 vậy
Bạn bảotự chứng minh được à, tại saolại như thế vậy ??!!
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ta có ab( a^22+b^22)\le≤2( tự CM)
=> ( a^22+ b^22)\le≤2/ab
=> ( a^22+ b^22)/2\le≤1/ab
làm tương tự ta có ( c^22+d^22)/2\le≤1/cd
cộng vế tương ứng vế. Hết.
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